Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, a[10100], y[10100], initial[10100], initialY[10100];
int basis[31], e;
bool inBasis[10100];
bool w[31][31];
vector<pair<int, int>> ans;
vector<pair<int, int>> after_ans;
void add(int x, int y) {
  a[x] ^= a[y];
  ans.push_back(make_pair(x + 1, y + 1));
}
void swp(int x, int y) {
  if (x == y) return;
  ans.push_back(make_pair(x + 1, y + 1));
  ans.push_back(make_pair(y + 1, x + 1));
  ans.push_back(make_pair(x + 1, y + 1));
  swap(a[x], a[y]);
}
void addToBasis(int x) {
  int i = 31 - 1;
  while (((a[x] >> i) & 1) == 0) --i;
  basis[i] = x;
  for (int j = 31 - 1; j > i; --j)
    if (basis[j] != -1 && (a[basis[j]] >> i) & 1) add(basis[j], basis[i]);
}
void normalize(int x) {
  for (int i = 31 - 1; i >= 0; --i)
    if (basis[i] != -1 && (a[x] >> i) & 1) {
      add(x, basis[i]);
    }
}
bool makeY(int num, bool toAns) {
  int Y = y[num];
  for (int i = 31 - 1; i >= 0; --i)
    if (basis[i] != -1 && (Y >> i) & 1) {
      Y ^= a[basis[i]];
      if (toAns) add(num, basis[i]);
    }
  return Y == 0;
}
void read() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    initial[i + 1] = a[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> y[i];
    initialY[i + 1] = y[i];
  }
}
void gaub() {
  fill(basis, basis + 31, -1);
  for (int i = 0; i < n; ++i) {
    normalize(i);
    if (a[i] > 0) {
      addToBasis(i);
      inBasis[i] = true;
    }
  }
}
bool check() {
  for (int i = 0; i <= n; ++i)
    if (!makeY(i, !inBasis[i])) return false;
  return true;
}
void finish_him() {
  for (int i = 31 - 1; i >= 0; --i)
    if (basis[i] != -1) {
      int need = -1;
      for (int j = i; j >= 0; --j)
        if (basis[j] != -1 && (y[basis[j]] >> i) & 1) need = j;
      if (need == -1) {
        add(basis[i], basis[i]);
        int Y = y[basis[i]];
        for (int j = i - 1; j >= 0; --j)
          if (basis[j] != -1 && (Y >> j) & 1) {
            Y ^= a[basis[j]];
            add(basis[i], basis[j]);
          }
      } else {
        swp(basis[i], basis[need]);
        swap(basis[i], basis[need]);
        int Y = y[basis[i]];
        for (int j = i - 1; j >= 0; --j)
          if (basis[j] != -1 && (Y >> j) & 1) {
            Y ^= a[basis[j]];
            add(basis[i], basis[j]);
          }
        for (int j = i - 1; j >= 0; --j)
          if (basis[j] != -1 && (y[basis[j]] >> i) & 1) {
            y[basis[j]] ^= a[basis[i]];
            after_ans.push_back(make_pair(basis[j] + 1, basis[i] + 1));
          }
      }
    }
  reverse(after_ans.begin(), after_ans.end());
  cout << ans.size() + after_ans.size() << "\n";
  for (auto& x : ans) cout << x.first << " " << x.second << "\n";
  for (auto& x : after_ans) cout << x.first << " " << x.second << "\n";
}
void doit() {
  for (auto& x : ans) {
    int a = x.first;
    int b = x.second;
    initial[a] ^= initial[b];
  }
  for (auto& x : after_ans) {
    int a = x.first;
    int b = x.second;
    initial[a] ^= initial[b];
  }
  for (int i = 1; i <= n; ++i, cout << " ") cout << initial[i];
  cout << "!\n";
  for (int i = 1; i <= n; ++i, cout << " ") cout << initialY[i];
}
int main() {
  ios_base::sync_with_stdio(0);
  read();
  gaub();
  if (!check()) {
    cout << "-1\n";
    return 0;
  }
  finish_him();
  return 0;
}